Given the balanced equation representing a reaction: #2CO(g) + O_2(g) -> 2CO_2(g)#. What is the mole ratio of #CO(g)# to #CO_2(g)# in this reaction?

1 Answer
May 20, 2018

Is it not #1:1#...?

Explanation:

We could write...#CO(g) +1/2O_2 rarr CO_2(g)#, instead of ...

#2CO(g) +O_2 rarr 2CO_2(g)#

In either scenario both mass and charge are balanced, as required for a stoichiometric equation...and this is why teachers go to such great efforts to teach stoichiometry....and we could put in numbers to stress the mass balance of the equation....

#underbrace(CO(g) +1/2O_2)_"28 g + 16 g = 44 g" rarr underbrace(CO_2(g))_"44 g"#